4U Maths Question Thread

[RuiAce]

Heyy

Which methods do I use to integrate (×+2) / (the root of: ×^2 - 1)

And also the root of: 9 -  x^2

The second one should be screaming trig sub. Which I get is unpleasant, but still it should be obvious because of the \( \sqrt{a^2-x^2} \) form..

The former is just a usual substitution. As for \( \int \frac{1}{\sqrt{x^2-1}} \,dx\), if this question came from an HSC paper prior to 2016 then you could just use the table of standard integrals to write down the answer. Otherwise, you have no choice but to use a trig sub as well, because of the \( \sqrt{x^2-a^2} \) form. Subbing in \(x = \sec \theta\) is pretty disgusting though, admittedly.

[vikasarkalgud]

Hey Rui, got this question from a past paper. Not sure how to do. (attached to post)

ty for any help

[RuiAce]

Hey Rui, got this question from a past paper. Not sure how to do. (attached to post)

ty for any help

[bdobrin]

Hey Rui,

Im really stuck on this recurrence formulae question. Any help would be appreciated

"If I_n = integral (from pi/2 to 0) of sinⁿxcos²x dx for n >0, show that I_n = [(n-1)/(n+2)] I_n-2"

thanks

[RuiAce]

Hey Rui,

Im really stuck on this recurrence formulae question. Any help would be appreciated

"If I_n = integral (from pi/2 to 0) of sinⁿxcos²x dx for n >0, show that I_n = [(n-1)/(n+2)] I_n-2"

thanks

\begin{align*}I_n &= \int_0^{\pi/2} \sin^n x \cos^2 x \,dx\\ &= \int_0^{\pi/2} \left( \sin^{n-1}x \cos^2 x \right) \sin x \, dx\\ &= \underbrace{\left[ -\cos x \left( \sin^{n-1}x \cos^2 x \right) \right]_0^{\pi/2}}_{\text{this thing equals 0}} \\&\qquad+ \int_0^{\pi/2} \cos x \Big( \left( (n-1) \cos x \sin^{n-2}x \right) \cos^2 x - (2\cos x \sin x)\sin^{n-1}x \Big) dx \\ &= (n-1) \int_0^{\pi/2} \sin^{n-2}x \cos^2x \cos^2 x \,dx - 2\int_0^{\pi/2} \sin^{n}x\cos^2 x \,dx\\ &= (n-1) \int_0^{\pi/2} \sin^{n-2}x \cos^2x \left( 1 - \sin^2 x \right)dx - 2I_n \\ \therefore 3I_n &= (n-1) \int_0^{\pi/2} \sin^{n-2}x \cos^2 x \, dx - (n-1) \int_0^{\pi/2} \sin^n x \cos^2 x \, dx\\ &= (n-1) I_{n-2} - (n-1) I_n\\ \therefore (n+2) I_n &= (n-1) I_{n-2}\\ I_n &= \frac{n-1}{n+2}I_{n-2}\end{align*}
It may be slower, but it might be a lot easier to express \(I_n\) in terms of \(J_{n+2}\), where \(J_n = \int_0^{\pi/2} \sin^n x\,dx\), and then try working with that.

[bdobrin]

Hey Rui,

thanks for that - didn't realise how hard that question was

also, how did you know to change sinⁿx to sin^n-1sinx ?

[vikasarkalgud]

Hey Rui, back again. 4u curve sketching question here and I'd like to know how exactly they got the integral graph to approach 2 in the final part, as x approaches negative infinity. Also, I had my own kind of way of doing f(e^x), was wondering if there is method to it coz mine is kind of dodgy. Thankyou very much.

[RuiAce]

Hey Rui,

thanks for that - didn't realise how hard that question was

also, how did you know to change sinⁿx to sin^n-1sinx ?

A bit of backtracking: I realised at an early onset that \( I_n = \int_0^{\pi/2} \sin^n x\,dx - \int_0^{\pi/2} \sin^{n+2} x \, dx\). But then I ignored that observation and kept trying to find smart ways to do this problem and surely enough, I was getting nowhere. So I started going back to my first idea and thought about how to handle the reduction formula for \(J_n = \int_0^{\pi/2} \sin^n x\,dx \). I recalled that for that one, I always had to split it into \( \sin x \sin^{n-1} x\) regardless. So (albeit with some reluctance) I decided to do that for this one as well. Thankfully it worked.

Hey Rui, back again. 4u curve sketching question here and I'd like to know how exactly they got the integral graph to approach 2 in the final part, as x approaches negative infinity. Also, I had my own kind of way of doing f(e^x), was wondering if there is method to it coz mine is kind of dodgy. Thankyou very much.

I don't see at all why the integral graph should approach 2. In fact, I don't know what it should approach at all - the only thing I know is that it must approach something greater than 0.

I would've done \(f(e^x)\) the same way I did \( f(x^2)\) tbh, which is also dodgy. When the transformation is under the brackets I usually don't rely on a toolkit, but rather visualise what's going on in my head. In a way, I visualise a "stretching" animation of the original graph.

Although, some things that are useful to note:
\( \lim_{x\to -\infty} f(e^x) = f(0) \), which is equal to 2 in this case so we expect our graph to asymptotically approach 2 towards the left
As \(x\to \infty\), \(f(e^x) \) is really just tending towards \( \lim_{x\to \infty}f(x) \), so we expect it to asymptotically approach 0 towards the right as well

[itssona]

ruii hey
would you happen to have a really simple proof that explains the rotation of a rectangular hyperbola? also if the hyperbola is in the form x^2-y^2=a^2, then do we just use the regular x=a/e for directrix right?

[RuiAce]

ruii hey
would you happen to have a really simple proof that explains the rotation of a rectangular hyperbola? also if the hyperbola is in the form x^2-y^2=a^2, then do we just use the regular x=a/e for directrix right?

It’s a bit hidden because the post is old but I actually have talked about it here.

And yeah. In particular, \(e=\sqrt2\) in this case

[itssona]

It’s a bit hidden because the post is old but I actually have talked about it here.

And yeah. In particular, \(e=\sqrt2\) in this case

ah thank you!:)

[justwannawish]

Hey, I'm a bit confused on how you integrate this with t-formula:
1dx/(3+4sin(2x))

[RuiAce]

Hey, I'm a bit confused on how you integrate this with t-formula:
1dx/(3+4sin(2x))

This is really just a pointless grind and involves nothing clever, only heavily brute forced computations.
\begin{align*}\int \frac{dx}{3 + 4\sin 2x} &=\int \frac{ \frac{2}{1+t^2}}{3 + \frac{8t}{1+t^2}}\,dt\\ &= \int \frac{2}{3t^2 + 8t + 3}\,dt\\ &=\frac23 \int \frac{1}{\left(t + \frac43\right)^2 - \frac79}\,dt \\ &=\frac23 \int \frac{1}{\left(t + \frac{4+\sqrt7}{3} \right)\left(t + \frac{4-\sqrt7}{3} \right)}\,dt \\ &= \frac1{\sqrt7}\int \left( \frac{1}{t + \frac{4-\sqrt7}3}+\frac{1}{t - \frac{4+\sqrt7}3} \right)dt\\ &= \frac1{\sqrt7} \left( \ln \left|  t + \frac{4-\sqrt7}{3}\right| - \ln \left| t + \frac{4+\sqrt7}{3} \right|\right)+C\\ &= \frac1{\sqrt7} \ln \left| \frac{3\tan x + 4 - \sqrt7}{3 \tan x + 4 +\sqrt7} \right|+C\end{align*}

[clovvy]

[RuiAce]

Did you get this question off the meme? (It's final answer is actually quite disgusting.)